Question

Solve the following triangles:

13. A side and a diagonal of a parallelogram are 12 inches and 19 inches respectively. The angle between the diagonals, opposite the given side, is 124°. Find the length of the other diagonal and the length of the other side.

14. A lighthouse is 10 miles northeast of a dock. A ship leaves the dock at noon, and sails east at a speed of 12 miles an hour. At what time will it be 8 miles from the lighthouse?

15.A vertical pole 35 feet high, standing on sloping ground, is braced by a wire which extends from the top of the pole to a point on the ground 25 feet from the foot of the pole. If the pole. subtends an angle of 30° at the point where the wire reaches the ground, how long is the wire?

16.A tower 125 feet high stands on the side of a hill. At a point 240 feet from the foot of the tower, measured straight down the hill, the tower subtends an angle of 25°. What angle does the side of the hill make with the horizontal?

Answer 1

(13)

Let b$=$ Side adjacent to angle of $124°$

Using cosine rule

$12^2=9.5^2+b^2=2(b)(9.5) cos 124$

$b^2+10.62b-53.75=0$

$b=\frac{-10.62{^+_-}\sqrt{10.62^2+4(53.75)}}{2}$

$b=3.74$ or $-14.36$

Length of the diagonal $=3.74×2$

$=7.48inches$

(14)

Let b $=$ distance sailed.

Using cosine Rule

$8^2=b^2+10^2-2×10×b\>cos\>45$

$b^2-14.14b+36=0$

$b=\frac{14.14{^+_-}\sqrt{14.14^2-4(36)}}{2}$

$=3.329\>or\>10.813$

Time taken $=\frac{3.329}{12}×60=16'39''$

Or $\frac{10.813}{12}×60=54'4''$

Time can be$;$

$12:16'39''\>p.m$

$Or\>12:54'4''\>p.m$

(15)

Let b$=$ length of the wire

Using cosine rule

$35^2=b^2+25^2-2×b×25\>cos\>30$

$b^2-43.30b-600=0$

$b=\frac{43.30{^+_-}\sqrt{43.3^2-4(-600)}}{2}$

$=54.34\>or\>-11.04$

Length of the wire $=54.34feet$

(16)

From a triangle 240 feet opposite to $\empty°$ and 125 feet opposite to $25°$

$\therefore$ $\frac{240}{Sin{\empty}}=\frac{125}{Sin25}$

$Sin{\empty}=\frac{240\>Sin25}{125}$

${\empty}=54.24°$

Let $\theta$$=$ angle to horizontal

$\theta=90-(54.24+25)$

$=10.76°$